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Angle of view - Ellipse and Hyperbola (Posted on 2011-11-24) Difficulty: 3 of 5
Given the equation x2/9 + y2/4 = 1 find the set of all points from which the angle of view* of this ellipse is a right angle. What is the significance of this set of points?

Given the equation x2/9 - y2/4 = 1 find the set of all points from which the angle of view* of this hyperbola is a right angle. What is the significance of this set of points?

* i.e. displaying a right angle between the two tangents.

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

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Solution An asymptotic view (spoiler) | Comment 2 of 7 |
I agree with Bractal’s algebraic results and use his notation, with s = +1 or -1:

A line through P(X, Y) with gradient m has equation y = Y + m(x – X)

and any points of intersection with the conic section  x2/a2 + sy2/b2 = 1 will

satisfy the equation   x2/a2 + s[Y + m(x – X)]2/b2 = 1        which can be

written as a quadratic in x as follows:

x2(b2 + sm2a2) + x[2sma2(Y – mX)] + sa2(Y – mX)2 – a2b2= 0

For the line to be a tangent to the conic this equation has equal roots for x.

Thus   [2sma2(Y – mX)]2 = 4((b2 + sm2a2)[ sa2(Y – mX)2 – a2b2]

which simplifies to:          m2s(X2 – a2) – m(2sXY) + sY2 – b2 = 0

For the two possible tangents from P to be perpendicular, the product of the roots of this quadratic in m must be -1.

Thus     sY2 – b2 = -s(X2 – a2), giving        X2 + Y2 = a2 + sb2           (1)

For s = 1 this is a circle, centre (0, 0) and radius sqrt(a2 + b2)

For s = -1 this is, potentially, a circle, centre (0, 0), radius sqrt(a2 – b2), but

the original conic is a hyperbola with  two distinct branches and with

asymptotes y = (b/a)x and y = -(b/a)x dividing the plane into four regions.

This means that two tangents to the same branch can only meet at points within the region occupied by that branch.

Case (i)  If a < b then (1) has no solution and no possible points exist.

Case (ii)  If a >= b then possible points lie on the two arcs of the circle (1)

which lie in the regions |Y| < (b/a)|X| containing the branches.

However, it seems to me that, while it makes sense to use tangents to define the limits of what can be seen of a closed curve, in the case of a hyperbola viewed from a point ‘outside’ the asymptotes of a particular branch, the ‘edge’ of that branch would be in the direction of the point at infinity on the asymptote; in other words, in the direction parallel to the asymptote.
This means that the two circular arcs found above can be extended tangentially

into the regions |Y| > (b/a)|X| to produce four rays drawn on the parallelogram

bY = +/- [aX +/- sqrt(a4 – b4)] which meet in pairs at (0, +/- sqrt(a4 – b4)/b)

In our case (a =3, b = 2) we have arcs on X2 + Y2 = sqrt(5) with rays on

2Y = +/- [3X +/- sqrt(65)], which meet at (0, +/- sqrt(65)/2) from which

points both  branches of the hyperbola present a right angle.



  Posted by Harry on 2011-11-25 23:17:28
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